Let $n>1$ be a natural number. We call a binary sequence $(b_1,\ldots, b_n)\in \{0,1\}^n$ $rigid$ if it is not a proper power of a sequence of shorter length. So for example $(0,1,0,1) = (0,1)^2$ is not rigid while $(0,0,1,1)$ is rigid. Differently phrased, we have an action of the cyclic group of order $n$, $C_n$, on the set of all binary sequences of length $n$, and a sequence is rigid if and only if $C_n$ acts freely on it. If $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ is the prime factorisation of $n$ then using the inclusion-exclusion principle the number of rigid sequences can be shown to be $$\sum_{d| p_1p_2\ldots p_k} (-1)^{r(d)}2^{\frac{n}{d}}$$ where $r(d)$ is the number of prime divisors of $d$. The number of $C_n$-orbits is then the above expression dividied by $n$.
On the other hand, consider the number of primitive elements in $\mathbb{F}_{2^n}$. These are elements $x\in \mathbb{F}_{2^n}$ which generate the field $\mathbb{F}_{2^n}$ over $\mathbb{F}_2$. A similar calculation using the inclusion-exclusion principle then shows that the number of primitive elements is equal to the number of rigid sequences in $\{0,1\}^n$. The Galois-group of $\mathbb{F}_{2^n}/\mathbb{F}_2$, which is also $C_n$, acts on the primitive elements. The orbits are in one to one correspondence with irreducible polynomials of degree $n$ over $\mathbb{F}_2$.
Question: Is there a known bijection between the set of rigid sequences of length $n$ and the set of primitive elements in $\mathbb{F}_{2^n}$? Is there a bijection between the corresponding $C_n$-orbit sets?
